SPOJ - Living with Courage Segment Tree

Question

Here is the problem link - http://www.spoj.com/problems/COURAGE/

My code gives WA. I use seg[] to represent segment sum tree and seg2[] to represent segment min tree. I do sum(range)-min(range) to output count in a range as per the problem.

#include <iostream>
#include <string>
#include <climits>
using namespace std;
long long int apples[100020],seg[200022],seg2[200022];
long long int min (long long int a, long long int b){
    if(a<b) return a;
    return b;
}
void buildt(long long int start,long long int end,long long int node){    
    if(start == end){
       seg[node] = apples[start];   
       seg2[node] = apples[start];
   }
   else{
       long long int mid = (long long int) ((start + end) / 2);
        buildt( start, mid,(long long int)(2*node));
        buildt( (long long int)(mid+1), end,(long long int)((2*node)+1));
        seg[node] = seg[2*node] + seg[(2*node)+1];
        seg2[node] = min(seg2[2*node] , seg2[(2*node)+1]);
    }    
}
void update(long long int node,long long int start,long long int end,long long int idx, long long int val)
{
   if(start == end )
   {
       apples[idx] = (long long int) (apples[idx] +  val);
       if(apples[idx] < 0){
           seg[node] = 0;
           seg2[node] = 0;
       }
       else{
       seg[node]  = (long long int)(seg[node]+ val);
       seg2[node] = (long long int)(seg2[node] + val);
       }
    }
    else
    {
        long long int mid = (long long int) ((start + end) / 2);
        if(start <= idx && idx <= mid)        
            update( (long long int)(2*node), start, mid, idx, val);        
        else
           update((long long int)((2*node)+1), (long long int)(mid+1), end, idx, val);
        seg[node] = seg[2*node] + seg[(2*node)+1];
       seg2[node] = min(seg2[2*node] , seg2[(2*node)+1]);
   }
}
long long int querysum(long long int node, long long int start, long long int end,long long int l,long long int r)
{ 
   if(r < start || end < l)
   {
       return 0;
    }
  if(l <= start && end <= r)
      return seg[node];
  long long int mid = (long long int) ((start + end) / 2);
  long long int p1 = querysum((long long int)(2*node), start, mid, l, r);
  long long int p2 = querysum((long long int)((2*node)+1), (long long int)(mid+1), end, l, r);
  return (p1 + p2);
}
long long int querymin(long long int node, long long int start, long long int end, long long int l, long long int r)
{
   if(r < start || end < l)
   {
       return LLONG_MAX;
   }
   if(l <= start && end <= r)
      return seg2[node];

  long long int mid = (long long int) ((start + end) / 2);
  long long int p1 = querymin((long long int)(2*node), start, mid, l, r);
  long long int p2 = querymin((long long int)((2*node)+1), (long long int)(mid+1), end, l, r);
  return min(p1 , p2);
}
int main()
{
   long long int n;
   cin >> n;
   for(int i = 1; i <= n; i++)
      cin >> apples[i];
  buildt(1,n,1);
  int op;
  cin >> op;
      string c;
      getline(cin,c);
  while(op--){
      cin >> c;
       char a1[64];
       char a2[64];
       cin >> a1 >> a2;
       if(c[0] == 'C')           
            cout <<  querysum(1,1,n,(long long int)(atoll(a1)+1),(long long int)(atoll(a2)+1)) -  querymin(1,1,n,(long long int)(atoll(a1)+1),(long long int)(atoll(a2)+1))<<endl;        
        else 
          if(c[0] == 'G')
             update(1,1,n,(long long int)(atoll(a2)+1),(long long int)(atoll(a1)));
          else
             update(1,1,n,(long long int)(atoll(a2)+1),(long long int)(-1*(atoll(a1))));  

        }          
}

Answer 1

I don't know much about C++, I run your code on spoj c++(g++ 4.3.2) It gave me compilation error. Then I modified it according to my comfort, it gave me SIGSEGV.(It ran fine on ideone though).

And now about problem: When number of apple become negative (as courage eats apple more than available) you make number of apple on that particular node equal to 0(zero). here is your code part for this:

if(start == end )
   {
       apples[idx] = (long long int) (apples[idx] +  val);
       if(apples[idx] < 0){
           seg[node] = 0;
           seg2[node] = 0;
       }
       else{
       seg[node]  = (long long int)(seg[node]+ val);
       seg2[node] = (long long int)(seg2[node] + val);
       }
    }

you made seg[node]=seg2[node]=0, but I think you forgot to modify your apples[node] to 0. So I think correct logic would be:

if(start == end )
   {
       apples[idx] = (long long int) (apples[idx] +  val);
       if(apples[idx] < 0){
           apples[idx]=0;
           seg[node] = 0;
           seg2[node] = 0;
       }
       else{
       seg[node]  = (long long int)(seg[node]+ val);
       seg2[node] = (long long int)(seg2[node] + val);
       }
    }

I made my program on java and I submitted two type of solution:

1. When I didn't check whether the number of apples are positive or negative.
2. When I checked whether the number of apples are positive or negative.

Funny part is both got accepted. If you want to take a look at my solution here is the link => solution And here is how I modified your code:

    #include <iostream>
#include <string>
#include <climits>
using namespace std;
long long int apples[100020],seg[200022],seg2[200022];
long long int min (long long int a, long long int b){
    if(a<b) return a;
    return b;
}
void buildt(long long int start,long long int end,long long int node){
    if(start == end){
       seg[node] = apples[start];
       seg2[node] = apples[start];
   }
   else{
       long long int mid = (long long int) ((start + end) / 2);
        buildt( start, mid,(long long int)(2*node));
        buildt( (long long int)(mid+1), end,(long long int)((2*node)+1));
        seg[node] = seg[2*node] + seg[(2*node)+1];
        seg2[node] = min(seg2[2*node] , seg2[(2*node)+1]);
    }
}
void update(long long int node,long long int start,long long int end,long long int idx, long long int val)
{
   if(start == end )
   {
       apples[idx] = (long long int) (apples[idx] +  val);
       if(apples[idx] < 0){
           apples[idx]=0;
           seg[node] = 0;
           seg2[node] = 0;
       }
       else{
       seg[node]  = (long long int)(seg[node]+ val);
       seg2[node] = (long long int)(seg2[node] + val);
       }
    }
    else
    {
        long long int mid = (long long int) ((start + end) / 2);
        if(start <= idx && idx <= mid)
            update( (long long int)(2*node), start, mid, idx, val);
        else
           update((long long int)((2*node)+1), (long long int)(mid+1), end, idx, val);
        seg[node] = seg[2*node] + seg[(2*node)+1];
       seg2[node] = min(seg2[2*node] , seg2[(2*node)+1]);
   }
}
long long int querysum(long long int node, long long int start, long long int end,long long int l,long long int r)
{
   if(r < start || end < l)
   {
       return 0;
    }
  if(l <= start && end <= r)
      return seg[node];
  long long int mid = (long long int) ((start + end) / 2);
  long long int p1 = querysum((long long int)(2*node), start, mid, l, r);
  long long int p2 = querysum((long long int)((2*node)+1), (long long int)(mid+1), end, l, r);
  return (p1 + p2);
}
long long int querymin(long long int node, long long int start, long long int end, long long int l, long long int r)
{
   if(r < start || end < l)
   {
       return LLONG_MAX;
   }
   if(l <= start && end <= r)
      return seg2[node];

  long long int mid = (long long int) ((start + end) / 2);
  long long int p1 = querymin((long long int)(2*node), start, mid, l, r);
  long long int p2 = querymin((long long int)((2*node)+1), (long long int)(mid+1), end, l, r);
  return min(p1 , p2);
}
int main()
{
   long long int n;
   cin >> n;
   for(int i = 1; i <= n; i++)
      cin >> apples[i];
  buildt(1,n,1);
  int op;
  cin >> op;
      string c;
      getline(cin,c);
  while(op--){
      cin >> c;
       long long a1,a2;
       cin >> a1 >> a2;

       if(c[0] == 'C')
            cout <<  querysum(1,1,n,a1+1,a2+1) -  querymin(1,1,n,a1+1,a2+1)<<endl;
        else
          if(c[0] == 'G')
             update(1,1,n,a2+1,a1);
          else
             update(1,1,n,a2+1,0-a1);

        }
}